Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{-5z^2 + 50z}{5z^2 + 15z - 140} \div \dfrac{z^2 - 10z}{z^2 - 14z + 40} $
Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-5z^2 + 50z}{5z^2 + 15z - 140} \times \dfrac{z^2 - 14z + 40}{z^2 - 10z} $ First factor out any common factors. $r = \dfrac{-5z(z - 10)}{5(z^2 + 3z - 28)} \times \dfrac{z^2 - 14z + 40}{z(z - 10)} $ Then factor the quadratic expressions. $r = \dfrac {-5z(z - 10)} {5(z - 4)(z + 7)} \times \dfrac {(z - 4)(z - 10)} {z(z - 10)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {-5z(z - 10) \times (z - 4)(z - 10) } { 5(z - 4)(z + 7) \times z(z - 10)} $ $r = \dfrac {-5z(z - 4)(z - 10)(z - 10)} {5z(z - 4)(z + 7)(z - 10)} $ Notice that $(z - 4)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-5z\cancel{(z - 4)}(z - 10)(z - 10)} {5z\cancel{(z - 4)}(z + 7)(z - 10)} $ We are dividing by $z - 4$ , so $z - 4 \neq 0$ Therefore, $z \neq 4$ $r = \dfrac {-5z\cancel{(z - 4)}\cancel{(z - 10)}(z - 10)} {5z\cancel{(z - 4)}(z + 7)\cancel{(z - 10)}} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $r = \dfrac {-5z(z - 10)} {5z(z + 7)} $ $ r = \dfrac{-(z - 10)}{z + 7}; z \neq 4; z \neq 10 $